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------------------------------------------------ A light beam 8.0 m long is pivoted at its center. A mass of 24 kg is placed on one end of the beam. How far from the center of the beam should a second mass of 36 kg be placed in order to balance the beam?

Give your answer in meters to 3 significant figures.

Distance from the center of the beam: ________ m

To balance the beam, the moments on both sides of the pivot need to be equal. The moment of a force is given by the product of the force and the distance from the pivot. In this case, the moment of the 24 kg mass is given by 24 kg × (8.0 m/2) = 96 kg*m. To balance this, the moment of the 36 kg mass must be equal. Let the distance from the centre of the beam to the 36 kg mass be x. Then, the moment of the 36 kg mass is given by 36 kg × x. Since the moments are equal, we can set up the equation 96 kg*m = 36 kg × x and solve for x.

Dividing both sides of the equation by 36 kg gives x = 96 kg*m / 36 kg = 2.67 m. So, the distance from the centre of the beam to the second mass should be approximately 2.67 m to balance the beam.

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